For a planet having mass equal to mass of the earth, the radius is one fourth of radius of the earth. Then escape velocity for this planet will be

The escape velocity of a celestial body is the minimum speed required for an object to break free from its gravitational influence without any further propulsion. For a planet with a mass equal to that of Earth but with a radius that is one-fourth of Earth’s radius, the escape velocity is much higher than that of Earth.

This is due to the relation between mass and radius in gravitational physics. The gravitational pull an object experiences on the surface of the planet depends on both the mass of the planet and the distance from its center. As the radius is reduced to one-fourth, the gravitational force at the surface increases, and it needs a greater speed to escape the gravitational field of the planet.

This means that the escape velocity for this hypothetical planet is about 22.4 km/s, which is double the escape velocity of Earth, which is about 11.2 km/s. This means that an object launched from the surface of this planet must achieve a much higher speed to overcome the stronger gravitational attraction. Understanding this concept is crucial for space missions and exploring the dynamics of celestial bodies in our universe.